A cylinder is rolling down on an inclined plane of inclination 60°. It's acceleration during rolling down will be x/√3 m / s², where x= ________ (use g=10 m / s²).

Solution

Physics sir · 4 steps

Part of gravity's pull goes into spin, so a rolling solid cylinder gets only 2/3g sin θ.

1. Rolling acceleration. For a body rolling without slipping, friction supplies the torque that spins it up. This makes the acceleration down the incline smaller than g sin θ.

   a = (g sin θ)/(1 + k²/R²)

2. Solid cylinder. Take the cylinder as solid, so I = 1/2MR² and k²/R² = 1/2. The denominator becomes 1 + 1/2 = 3/2.
3. Put in numbers. a = 2/3 × 10 × sin 60° = 2/3 × 10 × √3/2 = 10√3/3 m s⁻².
4. So. 10√3/3 = 10/√3 m s⁻². This matches x/√3, so x = 10.

The common slip: 15. Ignoring rotation and using a = g sin 60° = 5√3 = 15/√3 gives 15. That is the sliding answer, not the rolling one.

The idea to keep. For anything rolling down an incline, a = (g sin θ)/(1 + k²/R²). Put in k²/R²: 1/2 for a solid cylinder, 2/5 for a solid sphere, 1 for a ring.

Answer from the NTA final answer key. Solution by PrepWiser.